Question 941799
  equation of circle in standard form:  {{{(x - h)^2  +  (y-k)^2   =   r^2}}}
 where center  =(h,k) , radius  =r 
 center lies at the intersection of the lines 3x+y=1 and -2x-3y=4
          3x+y =1......eq(1)
   -2x-3y =4........eq(2)
*[invoke linear_substitution "x", "y", 3, 1, 1, -2, -3, 4 ]
(x,y)=(1,-2)
center(h,l)= (1,-2)
 equation circle becomes 
   {{{ (x-1)^2+(y-(-2))^2=6^2}}}
    {{{ (x-1)^2+(y+2)^2=36}}}
 Result : {{{ (x-1)^2+(y+2)^2=36}}}